Beauty of Array ZOJ - 3872(思维题)

 

Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the array A.

Input

There are multiple test cases. The first line of input contains an integer Tindicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 100000), which indicates the size of the array. The next line contains N positive integers separated by spaces. Every integer is no larger than 1000000.

Output

For each case, print the answer in one line.

Sample Input

3
5
1 2 3 4 5
3
2 3 3
4
2 3 3 2

Sample Output

105
21
38


题意就是给你一个数组,让你求1~1,1~2,1~3,1~4.... 1~n,
               2~2,2~3, 2~4.... 2~n
                 ....................
                      n-1~n-1 n-1~n
                           n~n
这些区间内不同数字的和,然后求总和。

思路:就是从右往左遍历,找某个数右边第一次出现这个数的位置,如果是第一次出现那么这个位置就是n
然后加加乘乘就行了


代码如下:
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 using namespace std;
 5 const int N=1e6+10;
 6 const int NN=1e5+100;
 7 int pos[N],a[NN];
 8 int main()
 9 {
10     int t;
11     int n;
12     scanf("%d",&t);
13     long long ans,tans;
14     while(t--)
15     {
16         scanf("%d",&n);
17         for(int i=0;i<n;i++)
18         {
19             scanf("%d",&a[i]);
20             pos[a[i]]=-1;//标记每个数都没出现过
21         }
22         ans=0,tans=0;
23         for(int i=n-1;i>=0;i--)
24         {
25             if(pos[a[i]]==-1)
26                 tans+=a[i]*(n-i);
27             else
28                 tans+=a[i]*(pos[a[i]]-i);
29             pos[a[i]]=i;
30             ans+=tans;
31         }
32         printf("%lld\n",ans);
33     } 
34     return 0;
35 }

 

posted @ 2018-11-29 21:04  甘雨说晚安吗  阅读(196)  评论(0编辑  收藏  举报
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